Stellar radii — practice questions

1. Convert distance to metres: $d = 100 \text{ ly} \times 9.46 \times 10^{15} \text{ m/ly} = 9.46 \times 10^{17} \text{ m}$
2. Calculate Luminosity (L) using the inverse square law for flux: $L = F \times 4\pi d^2$ $L = (1.50 \times 10^{-9} \text{ W m}^{-2}) \times 4\pi (9.46 \times 10^{17} \text{ m})^2$ $L = 1.68 \times 10^{28} \text{ W}$
3. Convert peak wavelength to metres and calculate Temperature (T) using Wien's Law: $λ_{max} = 450 \text{ nm} = 450 \times 10^{-9} \text{ m}$ $T = \frac{\text{constant}}{λ_{max}} = \frac{2.9 \times 10^{-3} \text{ m K}}{450 \times 10^{-9} \text{ m}}$ $T = 6444 \text{ K}$ (approx $6400 \text{ K}$)
4. Calculate Radius (r) using the Stefan-Boltzmann Law: Rearrange $L = 4\pi r^2 \sigma T^4$ to solve for r: $r = \sqrt{\frac{L}{4\pi \sigma T^4}}$ $r = \sqrt{\frac{1.68 \times 10^{28} \text{ W}}{4\pi (5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}) (6444 \text{ K})^4}}$ $r = \sqrt{\frac{1.68 \times 10^{28}}{4\pi (5.67 \times 10^{-8}) (1.73 \times 10^{15})}}$ $r = \sqrt{\frac{1.68 \times 10^{28}}{1.23 \times 10^9}}$ $r = \sqrt{1.37 \times 10^{19}}$ $r = 3.70 \times 10^9 \text{ m}$

The star's radius is approximately $3.70 \times 10^9 \text{ m}$ (or $3.7 \times 10^6 \text{ km}$).

1. State the Stefan-Boltzmann Law for both stars: For Betelgeuse (B): $L_B = 4\pi r_B^2 \sigma T_B^4$ For the Sun (S): $L_S = 4\pi r_S^2 \sigma T_S^4$

2. Create a ratio of the two equations to eliminate constants: Divide the equation for Betelgeuse by the equation for the Sun: $\frac{L_B}{L_S} = \frac{4\pi r_B^2 \sigma T_B^4}{4\pi r_S^2 \sigma T_S^4}$ The constants $4\pi$ and $\sigma$ cancel out: $\frac{L_B}{L_S} = \frac{r_B^2 T_B^4}{r_S^2 T_S^4} = (\frac{r_B}{r_S})^2 (\frac{T_B}{T_S})^4$

3. Rearrange the formula to solve for the radius ratio ($r_B/r_S$): $(\frac{r_B}{r_S})^2 = \frac{L_B}{L_S} \times (\frac{T_S}{T_B})^4$ $\frac{r_B}{r_S} = \sqrt{\frac{L_B}{L_S} \times (\frac{T_S}{T_B})^4}$

4. Substitute the given values and calculate: Luminosity ratio: $\frac{L_B}{L_S} = \frac{5.0 \times 10^{31} \text{ W}}{3.8 \times 10^{26} \text{ W}} = 1.316 \times 10^5$ Temperature ratio: $\frac{T_S}{T_B} = \frac{5800 \text{ K}}{3500 \text{ K}} = 1.657$

   Now substitute these into the rearranged formula: $\frac{r_B}{r_S} = \sqrt{(1.316 \times 10^5) \times (1.657)^4}$ $\frac{r_B}{r_S} = \sqrt{(1.316 \times 10^5) \times 7.553}$ $\frac{r_B}{r_S} = \sqrt{9.936 \times 10^5}$ $\frac{r_B}{r_S} = 996.8$

Betelgeuse's radius is approximately 997 times larger than the Sun's radius. This demonstrates why it is classified as a supergiant: it is extremely large to be so luminous despite its relatively cool surface temperature.